If   \vec{a}=\frac{1}{\sqrt{10}}\left ( 3\hat{i}+\hat{k} \right )\; and\;\; \vec{b}=\frac{1}{7}\left ( 2\hat{i}+3\hat{j}-6\hat{k} \right ),   then the value of (2\vec{a}-\vec{b})\cdot \left [ \left (\vec{a}\times \vec{b} \right )\times \left ( \vec{a}+2\vec{b} \right ) \right ]\; \; is

  • Option 1)

    5

  • Option 2)

    3

  • Option 3)

    -5

  • Option 4)

    -3

 

Answers (1)

As we learnt in 

Vector Product of two vectors(cross product) -

If \vec{a} and \vec{b} are two vectors and \Theta is the angle between them , then \vec{a}\times \vec{b}=\left |\vec{a} \left | \right |\vec{b} \right |Sin\Theta \hat{n}

- wherein

\hat{n} is unit vector perpendicular to both \vec{a} \: and \: \vec{b}

 

 

Scalar Triple Product -

\left [ \vec{a}\;\vec{b}\; \vec{c} \right ]

=\left (\vec{a}\times \vec{b}\right)\cdot \vec{c}= \vec{a}\cdot \left ( \vec{b} \times \vec{c}\right )

=\left (\vec{b}\times \vec{c}\right)\cdot \vec{a}= \vec{b}\cdot \left ( \vec{c} \times \vec{a}\right )

=\left (\vec{c}\times \vec{a}\right)\cdot \vec{b}= \vec{c}\cdot \left ( \vec{a} \times \vec{b}\right )

- wherein

Scalar Triple Product of three vectors \hat{a},\hat{b},\hat{c}.

 

 \vec{a}=\frac{1}{\sqrt{10}}(3\hat{i}+\hat{k})

\vec{b}=\frac{1}{7}(2\hat{i}+3\hat{j}-6\hat{k})

\left | \vec{a} \right |=\left | \vec{b} \right |=1;\:\vec{a}\:.\vec{b}\:=0

\left | \vec{a}\times \vec{b} \right |=\left | \vec{a} \right |\:\left | \vec{b} \right |\:sin90^{\circ}\:=1

STP [2\vec{a}-\vec{b}\:\:\:\:\vec{a}\times \vec{b}\:\:\:\:\:\vec{a}+2\vec{b}]

=(\vec{a}\times \vec{b})\:.\:[(\vec{a}+2\vec{b})\times(2\vec{a}-\vec{b})]

=(\vec{a}\times\vec{b})\:.\:5(\vec{b}\times\vec{a})=-5\:(\vec{a}\times \vec{b})^{2}

We saw (\vec{a}\times \vec{b})=1

So value = -5


Option 1)

5

This option is incorrect.

Option 2)

3

This option is incorrect.

Option 3)

-5

This option is correct.

Option 4)

-3

This option is incorrect.

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